circle geometry question (1 Viewer)

[Doherty]

ABC is a triangle in which BC=AC. D is a point on the minor arc AC of the cirlce passing through A,B,C. AD is produced to E
i)copy the diagram
ii)Give a reason why LCDE=LABC
iii) Hence show that DC bisects LBDE

i have done i and ii but i just cant think of how to do iii

 

[lollol]

i think if u use the theorem "angles in the same segment are equal", then u can get LBAC=LBDC, so LABC=LBDC, LCDE=LBDC

 

[ianc]

so LABC=LBDC, LCDE=LBDC

this last bit uses the cyclic quad (exterior angle = interior opposite angle) rule

 

[SoulSearcher]

To fully work this out:
@CDE = @CBA (exterior angle = interior opposite angle)
AC = BC (given)
So triangle ABC is isosceles
Therefore,
@CAB = @CBA (base angles in a isos. triangle are equal)
but
@CAB = @CDB (angles in the same segment are equal)
Therefore
@CDB = @ CDA
and
@CDB = @CDE
Therefore DC bisects @BDE

 

[*slurtznow&4eva]

ive got an assignment due tomorrow for extension please help :

AB & CD are two intersecting chords of a circle and CD is parallel to the tangent to the circle at B. prove that AB bisects angle CAD
please help

 

[SoulSearcher]

@FBD = @BDC (alternate angles, CD || EF)
but @FBD = @DAB (alternate segment theorem)
and @CAB = @BDC (angles in the same segment)
.'. @CAB = @DAB
Therefore AB bisects @CAD